For the case \(0<\alpha <n,\)  the boundedness of the fractional maximal
operator \(M_{\alpha }\)  from \(L^{p}(w^{p})\)  into \(L^{q}(w^{q})\)  for \(1<p<\frac{n}{\alpha }\) , \(\frac{1}{q}=\frac{1}{p}-\frac{\alpha }{n}\)  \(\ \) and for
all weights \(w\)  in the Muckenhoupt class \(A_{p,q}\)  (see [1]}) gives
the inequality 3.16, Theorem 3.23 in [2]}, for the pair \(\left(M_{\alpha }f,f\right) \) , \(f\in L^{p}\left( \mathbb {R}^{n}\right) .\)  So for \(1<\theta <\infty ,\)
\(\left\Vert \left\lbrace  \sum \limits _{j=1}^{\infty }\left( M_{\alpha }f_{j}\right)^{\theta }\right\rbrace  ^{\frac{1}{\theta }}\right\Vert _{L^{q}\left(w^{q}\right) }\lesssim \left\Vert \left( \sum \limits _{j=1}^{\infty }\left|f_{j}\right|^{\theta }\right) ^{\frac{1}{\theta }}\right\Vert _{L^{p}\left( w^{p}\right) }\) 
for all weights \(w\)  in the Muckenhoupt class \(A_{p,q}.\)  Now \(w\in A_{1}\) 
implies \(w^{\frac{1}{q}}\in A_{p,q}\)  so
\(\left\Vert \left\lbrace  \sum \limits _{j=1}^{\infty }\left( M_{\alpha }f_{j}\right)^{\theta }\right\rbrace  ^{\frac{1}{\theta }}\right\Vert _{L^{q}\left( w\right)}\lesssim \left\Vert \left( \sum \limits _{j=1}^{\infty }\left|f_{j}\right|^{\theta }\right) ^{\frac{1}{\theta }}\right\Vert _{L^{p}\left( w^{\frac{p}{q}}\right) }\) 
for all \(w\in A_{1},\)  thus the lemma follows , in this case, from Lemma
4.30 in [2]}. For the case \(\alpha =0,\)  Theorem 4.25 in [2]}
applies.
